I usually don't have any problem with these, but this one has me stuck.

1 - sinA = (sin(A/2) - cos(A/2))^(2)

subbing in the half-angle identities

1 - sinA = (((1 - cosA) / 2)^(1/2) - ((1 + cosA) / 2)^(1/2))^(2)

adding the complex fractions to simplify

1 - sinA = (((1 - cosA)^(1/2) - (1 + cosA)^(1/2)) / 2)^(2)

squaring (FOIL)

1 - sinA = ((1 - cosA) - (1 - cosA) - (1 - cosA) + (1 - cosA))/2

simplify

1 - sinA = 2cosA/2

not true

1 - sinA = cosA

It is tempting to say cos^(2)A = 1 - sin^(2)A, so cosA = 1 - sinA, but last time I checked, it does not work like that. CosA = (1 - sin^(2)A)^(1/2), and since that is a false square -- (1 - sinA)(1 + sinA) -- there is no real way to take the root. Tell me where I am wrong.